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Rashid [163]
3 years ago
14

Determine the volume, in liters, occupied by 0.015 molecules of oxygen at STp?

Chemistry
1 answer:
Arlecino [84]3 years ago
5 0

5.58 X 10^{-25} Litres is the volume, in liters, occupied by 0.015 molecules of oxygen at STP.

Explanation:

Data given:

molecules of oxygen = 0.015

number of moles of oxygen =?

temperature at STP = 273  K

Pressure at STP = 1 atm

volume = ?

R (gas constant) = 0.08201 L atm/mole K

to convert molecules to moles,

number of moles = \frac{molecules}{Avagadro's number}

number of moles = 2.49 x 10^{-26}

Applying the ideal gas law since the oxygen is at STP,

PV = nRT

rearranging the equation:

V = \frac{nRT}{P}

putting the values in the rearranged equation:

V = \frac{2.49 X10^{-26}   X 0.08201 X 273}{1}

V = 5.58 X 10^{-25} Litres.

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The following calculations must be handwritten in your notebook. – Acetic Acid ■ Hydrogen ion concentration ■ Ka ■ % Error – Ace
Nikitich [7]

Answer:

Explanation:

1) Acetic acid

Concentration is given as 0.103 M

The average pH of this solution = 2.96

we know that pH = - log [H+] therefore [H+] = 10-pH

[H+] = 10-2.96

= 1.1 x 10-3 M = 0.0011 M

Consider the equilibrium

CH3COOH ⇄CH3COO- + H+

Initial 0.103 0 0

Change -x +x +x

equlibrium 0.103 -x x x

Ka = x2 / 0.103 - x

Here the initial concentration of CH3COOH = 0.103 M

the equilibrium concentration of H+ = x = 0.0011 M

Therefore the equilibrium conc of acetic acid = 0.103 - 0.0011 = 0.1019 M

Therefore Ka = 0.0011 x 0.0011 / 0.1019 = 1.187 x 10-5

2) Acetic acid + NaOH

pH measured = 4.48 , therefore [H+} = 10-4.48 = 3.3 x 10-5

Volume and conc of acetic acid = 10 mL of 0.103 M

= 10 mL x 0.103 mmol / mL

= 1.03 mmol

Volume and conc of NaOH added = 4 mL of 0.0992 M

= 4 x 0.0992 mmol

= 0.397 mmol

Consider the equation

CH3COOH + NaOH -----------> CH3COONa + H2O

Initial 1.03 0.397 0

Final 0.633 0 0.397

0.397 mmole of NaOH will convert 0.397 mmole of acetic acid to sodium acetate.

Thus the final moles of acetic acid and sodium acetate in the solution are 0.633 and 0.397

therefore [salt] / [acid] = 0.397 / 0.633 = 0.627

By Hendersen equation pH = pKa + log[salt / acid]

pH = pKa + log 0.627 = pKa - 0.203

or pKa = pH + 0.203 = 4.48 + 0.203 [ since the measured pH = 4.48]

= 4.683

Ka = 10-4.683 = 2.07 x 10-5

3) Phosphate salts:

(i) mass of NaH2PO4 taken = 0.613 g

molar mass of NaH2PO4 = 120

therefore moles = 0.613 / 120 = 0.0051 mole

= 5.1 mmol

The volume is 30 mL therefore concentration = 5.1 /30 mmol/mL

= 0.17 M

consider the equilibrium

H2PO4-⇄ HPO42- + H+

Initial 0.17 0 0

Change -x +x + x

equilibrium 0.17-x x x

Ka = x2 / 0.17-x = 6.2 x 10-8 [ Ka is given]

neglect x in the denominator as it is very small x2 = 0.17 x 6.2 x 10-8

x =  1.03 x 10-4

Thus the equilirium conc of H+ = 1.03 x 10-4 therefore pH = - log 1.03 x 10-4 = 3.99

(ii) Mass of Na2HPO4.7H2O =0.601 g

therefore no of moles = 0.601 / 268.07 = 0.00224 mole

= 2.24 mmol

The volume = 30 mL , therefore conc = 2.24 / 30 mmol/ml

= 0.075 M

consider the equilibrium

HPO42- ⇄ PO43- + H+

Initial 0.075 0 0

Change -x +x + x

equilibrium 0.075-x x x

Ka = x2 / 0.075-x = 4.8 x 10-13 [ Ka is given]

neglect x in the denominator as it is very small x2 = 0.075 x 4.8 x 10-13

x =  1.9 x 10-7

Thus the equilirium conce of H+ = 1.9 x 10-7 therefore pH = - log 1.9 x 10-7 = 6.7

(iii) Mass of Na3PO4.12H2O taken = 0.208 g

moles of trisodiumphosphate 0.208/ 380 = 0.00055 moles

= 0.55 mmol

Volume = 10 mL therefore conc = 0.55/10 = 0.055 mmol/mL

= 0.055 M

Consider the equilibrium reaction

PO43- + H2O  ⇄ HPO42- + OH-

initial 0.055 0 0

Change -x +x +x

equilibrium 0.055-x x x

Kb = x2/ 0.055 -x = 0.0208 [Kb = Kw / Ka = 10-14 / 4.8 x 10-13 = 0.0208]

x2 + 0.0208x - 0.001144 = 0 Solving this equation we get x = 0.025

That is the conce of OH- ion = 0.025M

Therefore pH = 14 - pOH = 14 - 1.6 =12.4

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2 years ago
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Answer:

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Explanation:

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3 years ago
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Which particle changes to create an isotope of an atom? <br> A.proton<br> B.neutron<br> C.electron
Dovator [93]

Answer:

A

Explanation:

proton

5 0
3 years ago
How many electrons does an element have if its atomic number is 20
olga55 [171]

Answer:

20 electrons

A neutral atom with atomic number 20 will have 20 electrons.

The atomic number is, by definiton, the number of protons in an atom's nucleus but for a neutral atom it's also equal to the number of electrons. Each element has a different, unique number of protons that determines its identity.

Explanation:

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2 years ago
Which buffer would be better able to hold a steady pH on the addition of strong acid, buffer 1 or buffer 2? Explain. Buffer 1: a
Talja [164]

Answer:

Buffer 1.

Explanation:

Ammonia is a weak base. It acts like a Bronsted-Lowry Base when it reacts with hydrogen ions.

\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq).

\rm NH_3 gains one hydrogen ion to produce the ammonium ion \rm {NH_4}^{+}. In other words, \rm {NH_4}^{+} is the conjugate acid of the weak base \rm NH_3.

Both buffer 1 and 2 include

  • the weak base ammonia \rm NH_3, and
  • the conjugate acid of the weak base \rm {NH_4}^{+}.

The ammonia \rm NH_3 in the solution will react with hydrogen ions as they are added to the solution:

\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq).

There are more \rm NH_3 in the buffer 1 than in buffer 2. It will take more strong acid to react with the majority of \rm NH_3 in the solution. Conversely, the pH of buffer 1 will be more steady than that in buffer 2 when the same amount of acid has been added.

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